Lecture Notes on Graduate Electrodynamics
2016
1 Review of Electrostatics
The first 11 sections (you can omit 1.12 and 1.13) of this chapter should only be revision for you; little is done there that
you shouldn’t have seen before (presumably...). Therefore, I will assign it for you to read carefully, and will only go over the
salient points here.
1.1
Electrostatic Field (sections J1.1–J1.4)
Coulomb’s law gives the force exerted on a point-charge q2 by a point-charge q1 :
F = k e q1 q2
x2 − x1
|x2 − x1 |3
(1.1)
where x2 − x1 is the distance vector from the source charge q1 , located at position x1 , to q2 , located at x2 , with respect to
an arbitrarily chosen origin. In SI units, ke = 1/4πǫ0 = 8.99 × 109 N · m2 /C2 ; in CGS units, it is simply equal to 1. To
avoid any confusion I will adopt SI units as is done in the first ten chapters of Jackson.
We also define the Coulomb electric field whose source is q as:
E(x) = ke q
x − x′
|x − x′ |3
(1.2)
Again, x − x′ is the distance vector from the source at x′ to the observation (or field) point x, where there may or may not
exist a point-charge (see fig. J1.1). So F = qE.
The electric field obeys the principle of superposition. So, when the source is a collection of N point-charges qi sitting
at their respective positions x′i , the electric field at observation point x is:
E(x) = ke
N
X
i=1
qi
x − x′ i
|x − x′ i |3
This is easily extended to continuous source charge distributions. We only give the expression for a distribution extending
over a volume:
Z
x − x′ 3 ′
E(x) = ke
d x
(1.3)
ρ(x′ )
|x − x′ |3
volume
where ρ is the charge density in the distribution. Note that only Cartesian components can be used in the integral, because
only then are the unit vectors in x − x′ constant with respect to integration, so that they can be pulled out of the integral. The
remaining three integrals can then be evaluated in any coordinate basis you wish.
Note also that you can always write a discrete distribution of N point-charges as:
ρ(x) =
N
X
qi δ(x − xi )
i=1
The Coulomb field, as well as any electric field, satisfies Gauss’ law in differential form:
∇ · E = 4πke ρ
or, in the equivalent integral form:
where qin =
R
I
(= ρ/ǫ0 in SI units)
E · da = 4πke qin
(1.4)
(1.5)
surface
ρ(x′ ) d3 x′ is the total net charge enclosed by the surface.
The field of any static charge distribution can be written as a superposition of Coulomb fields. We call such fields
electrostatic.
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Lecture Notes on Graduate Electrodynamics
1.2
2016
Scalar Potential (sections J1.5–J1.7)
Because it is a central field, the Coulomb field has vanishing curl. Therefore,
by the superposition principle, all electrostatic
H
fields satisfy ∇ × E = 0 everywhere. This is equivalent to saying that E · dl = 0 for any closed integration path.
Because ∇ × E = 0 for any electrostatic field E, there exists associated with this vector field a scalar field Φ(x) such
that E = −∇Φ. From the fundamental theorem for gradients (section 0.3):
Z
b
E · dl = −
a
Z
b
(∇Φ) · dl = Φ(a) − Φ(b)
(1.6)
a
Therefore, the electrostatic potential at any point P is Φ(P ) = −
chosen where the potential is zero.
RP
O
E · dl, where O is an arbitrary reference point, usually
The Coulomb potential of a point-charge q is Φ(x) = ke q/|x − x′ |.
The potential Φ, like the field E, obeys the superposition principle. Then the potential of a localised (ie. with compact
support) charge distribution ρ is:
Z
ρ(x′ ) 3 ′
Φ(x) = ke
d x
(1.7)
|x − x′ |
where the reference point is taken at infinity, hence the restriction to distributions with compact support.
Combining E = −∇ Φ with Gauss’ law yields the Poisson equation:
∇2 Φ = − 4πke ρ
(1.8)
In vacuum, this becomes the Laplace equation, ∇2 Φ = 0.
1.3
Matching Conditions on Electrostatic E and Φ
Let n̂ be a unit vector normal to the surface at some point, directed from the “in” region to the “ out” region. The electrostaticfield component tangent to any surface is continuous across the surface:
(Eout − Ein ) × n̂ = 0
(1.9)
For conductors, on the other hand, the normal component is discontinuous when there is a local surface charge density σ:
(Eout − Ein ) · n̂ = = 4πke σ
(1.10)
The local field due to σ has magnitude 2πke σ, so that the total field has to jump by σ/4πke across the surface.
The potential is continuous across the boundary, but its gradient in the direction normal to the surface inherits the discontinuity in E:
∂n Φout − ∂n Φin = − 4πke σ
(1.11)
where the normal derivative , ∂n Φ = (∇ Φ) · n̂, is evaluated at the surface.
For non-conductors, we will have to be a bit careful (see chapter 4 in Jackson) with the normal components.
1.4
Boundary-Value Problem with Green Functions; Uniqueness of a Solution (sections J1.9–
J1.10)
In realistic electrostatic problems, we rarely know the whole charge distribution that produces the potential observed at a
point. What we most often do know (and often can control) is the potential over specific surfaces such as conductors. These
surfaces can enclose regions with or without charge inside. The proper approach then involves solving the Poisson or Laplace
equation, subject to boundary conditions (B.C.).
There are two very important questions that must be answered:
• What kind of information do we need to supply in order to obtain a solution to the Poisson equation?
• If we find a solution, is it unique?
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